A train is moving at a speed of 50 km/h. How many hours will it take thetrain to travel 600 kilometers

Debby and Ben took different routes to travel from Point A to Point E. Debby took the route along A, B, C, D, and E. Ben took th

svetlana [45]

Answer:

Ben's average speed was twice Debby's average speed.

Explanation:

Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.

4 0

3 years ago

Read 2 more answers

What happens to the velocity of a sound wave in air if the temperature of the air increases?

Yuki888 [10]

The answer is D :) hope this helps

6 0

3 years ago

Read 2 more answers

When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the

MaRussiya [10]

Answer:

(a) $\alpha=-111.26rad/s$

(b) $s=4450.6in$

(c) $8.66in$

Explanation:

First change the units of the velocity, using these equivalents $1rev=2\pi rad$ and $1 min =60s$

$4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s$

The angular acceleration $\alpha$ the time rate of change of the angular speed $\omega$ according to:

$\alpha=\frac{\Delta \omega}{\Delta t}$

$\Delta \omega=\omega_i-\omega_f$

Where $\omega_i$ is the original velocity, in the case the velocity before starting the deceleration, and $\omega_f$ is the final velocity, equal to zero because it has stopped.

$\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s$

b) To find the distance traveled in radians use the formula:

$\theta = \omega_i t + \frac{1}{2} \alpha t^2$

$\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad$

To change this result to inches, solve the angular displacement $\theta$ for the distance traveled $s$ ( $r$ is the radius).

$\theta=\frac{s}{r} \\s=\theta r$

$s=890.12(5)=4450.6in$

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

$\frac{890.12}{2\pi}=141.6667$

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle $\gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120$ is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which is also the net displacement):

$c^2=a^2+b^2-2abcos(\gamma)$

$c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in$

4 0

4 years ago

A high school physics student is sitting in a seat read-

Nataly_w [17]

The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

$N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}$

Let's analyze the forces on the student.

$N_{student} - W_{student} = 0 \\N_{student} = W _{student}$

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Learn more here: brainly.com/question/18117041

7 0

3 years ago

A girl drops a stone from the top a tower 45m tall. At the same time, a boy standing at the base of the tower, projects another

Advocard [28]

Answer:

(i) The stones meet at 1.8 second

(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

Explanation:

(i)

First we consider the stone dropped by the girl. We have data:

Vi = Initial Velocity of Stone = 0 m/s (Since the stone was initially at rest)

t = Time Period

g = 10 m/s²

s₁ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₁ = Vi t + (0.5)gt²

s₁ = (0)(t) + (0.5)(10)t²

s₁ = 5t² ----- equation (1)

Now, we consider the stone throne vertically upward by the boy. We have data:

Vi = Initial Velocity of Stone = 25 m/s

t = Time Period

g = - 10 m/s² (negative sign due to upward motion)

s₂ = Distance Covered by Stone

Using 2nd equation of motion, we get:

s₂ = Vi t + (0.5)gt²

s₂ = (25)(t) + (0.5)(-10)t²

s₂ = 25t - 5t² ----- equation (2)

At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).

s₁ + s₂ = 45

using values from equation (1) and equation (2)

5t² + 25t - 5t² = 45

25t = 45

t = 45/25

t = 1.8 sec

(ii)

using this value of of t in equation (2)

s₂ = (25)(1.8) - (5)(1.8)²

s₂ = 28.8 m

using this value of of t in equation (1)

s₁ = (5)(1.8)²

s₁ = 16.2 m

Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.

6 0

3 years ago